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Ex 9.3, 12 Find a particular solution satisfying the given condition : ๐‘ฅ(๐‘ฅ^2โˆ’1) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=1;๐‘ฆ=0 When ๐‘ฅ=2 ๐‘ฅ(๐‘ฅ^2โˆ’1) dy = dx dy = ๐’…๐’™/(๐’™(๐’™๐Ÿ โˆ’ ๐Ÿ)) Integrating both sides. โˆซ1โ–’๐‘‘๐‘ฆ = โˆซ1โ–’๐‘‘๐‘ฅ/(๐‘ฅ(๐‘ฅ2 โˆ’ 1)) ๐’š = โˆซ1โ–’๐’…๐’™/(๐’™(๐’™ + ๐Ÿ)(๐’™ โˆ’ ๐Ÿ)) We can write integrand as ๐Ÿ/(๐’™(๐’™ + ๐Ÿ)(๐’™ โˆ’ ๐Ÿ)) = ๐‘จ/๐’™ + ๐’ƒ/(๐’™ + ๐Ÿ) + ๐’„/(๐’™ โˆ’ ๐Ÿ) By canceling the denominators. 1 = A (x โˆ’ 1) (x + 1) B x (x โˆ’ 1) + C x (x + 1) Putting x = 0 1 = A (0 โˆ’ 1) (0 + 1) + B.0. (0 โˆ’ 1) + C.0. (0 + 1) 1 = A (โˆ’1) (1) + B.0 + C.0 1 = โˆ’ A A = โˆ’1 Similarly putting x = โˆ’1 1 = A (โˆ’1 โˆ’ 1) (โˆ’1 + 1) + B (โˆ’1) (โˆ’1 โˆ’ 1) + C(โˆ’1)(โˆ’1 + 1) 1 = A (โˆ’2) (0) + B (โˆ’1) (โˆ’2) + C (โˆ’1) (0) 1 = 0 + 2B + 0 2B = 1 B = ๐Ÿ/๐Ÿ Similarly putting x = 1 1 = A(1 โˆ’ 1) (1 + 1) + B.1(1 โˆ’ 1) + C(1)(1 + 1) 1 = A (0) (2) + B.1.0 + C.2 2C = 1 C = ๐Ÿ/๐Ÿ Therefore, ๐Ÿ/(๐’™(๐’™ + ๐Ÿ)(๐’™ โˆ’ ๐Ÿ)) = (โˆ’๐Ÿ)/๐’™ + ๐Ÿ/(๐Ÿ(๐’™ + ๐Ÿ)) + ๐Ÿ/(๐Ÿ(๐’™ โˆ’ ๐Ÿ)) Now, From (1) y = โˆซ1โ–’1/(๐‘ฅ(๐‘ฅ + 1)(๐‘ฅ โˆ’ 1)) dx = โˆ’ โˆซ1โ–’๐Ÿ/๐’™ + dx + ๐Ÿ/๐Ÿ โˆซ1โ–’๐’…๐’™/(๐’™ + ๐Ÿ) + ๐Ÿ/๐Ÿ โˆซ1โ–’๐’…๐’™/(๐’™ โˆ’ ๐Ÿ) = log |๐’™|+ ๐Ÿ/๐Ÿ log |๐’™+๐Ÿ| + ๐Ÿ/๐Ÿ log |๐’™โˆ’๐Ÿ|+๐’„ = (โˆ’2)/2 logโก|๐‘ฅ| + ๐Ÿ/๐Ÿ log |๐’™+๐Ÿ|+ ๐Ÿ/๐Ÿ log |๐’™โˆ’๐Ÿ|+๐‘ = 1/2 [โˆ’2 logโกใ€–|๐‘ฅ|โˆ’2+๐ฅ๐จ๐ โก|(๐’™+๐Ÿ)(๐’™โˆ’๐Ÿ)| ใ€— ]+๐‘ = 1/2 [logโกใ€–๐‘ฅ^(โˆ’2)+logโก|(๐‘ฅ+1)(๐‘ฅโˆ’1)| ใ€— ]+๐‘ = 1/2 [logโก|๐‘ฅ^(โˆ’2) (๐‘ฅ^2โˆ’1)| ]+๐‘ = ๐Ÿ/๐Ÿ log |(๐’™^๐Ÿ โˆ’ ๐Ÿ)/๐’™^๐Ÿ |+๐’„ Given that x = 2, y = 0 Substituting values in (1) we get 0 = 1/2 " log " |(2^2โˆ’1)/2^2 |" + C" 0 = 1/2 " log " 3/4 " + C" C = โˆ’๐Ÿ/๐Ÿ " log " ๐Ÿ‘/๐Ÿ’ Putting value of c in (1), y = ๐Ÿ/๐Ÿ log |(๐’™^๐Ÿ โˆ’ ๐Ÿ)/๐’™^๐Ÿ | โˆ’ ๐Ÿ/๐Ÿ log ๐Ÿ‘/๐Ÿ’

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo